A244051 - OEIS

A244051

Triangle read by rows in which row n lists the parts of the partitions of n into equal parts, in nonincreasing order.

1, 2, 1, 1, 3, 1, 1, 1, 4, 2, 2, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 6, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 8, 4, 4, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 9, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 10, 5, 5, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
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	OFFSET	`1,2`
	COMMENTS	`Row n has length sigma(n) =A000203(n).` `Row sums give nA000005(n) =A038040(n).` `Column 1 isA000027.` `Both columns 2 and 3 areA032742,n > 1.` `For any k > 0 and t > 0, the sequence contains exactly one run of k consecutive t's. -Rémy Sigrist,Feb 11 2019` `FromOmar E. Pol,Dec 04 2019: (Start)` `The number of parts congruent to 0 (mod m) in row mn equals sigma(n) =A000203(n).` `The number of parts greater than 1 in row n equalsA001065(n), the sum of aliquot parts of n.` `The number of parts greater than 1 and less than n in row n equalsA048050(n), the sum of divisors of n except for 1 and n.` `The number of partitions in row n equalsA000005(n), the number of divisors of n.` `The number of partitions in row n with an odd number of parts equalsA001227(n).` `The sum of odd parts in row n equals the sum of parts of the partitions in row n that have an odd number of parts, and equals the sum of all parts in the partitions of n into consecutive parts, and equalsA245579(n) = n*A001227(n).` `The decreasing records in row n give the n-th row ofA056538.` `Row n has n 1's which are all at the end of the row.` `First n rows containA000217(n) 1's.` `The number of k's in row n isA126988(n,k).` `The number of odd parts in row n isA002131(n).` `The k-th block in row n hasA027750(n,k) parts.` `Right border givesA000012.(End)` `The r-th row of the triangle begins at index k =A160664(r-1). -Samuel Harkness,Jun 21 2022`
	LINKS	`Samuel Harkness,Table of n, a(n) for n = 1..10000` `Samuel Harkness,Log-log Scatterplot of the first 1000000 terms`
	EXAMPLE	`Triangle begins:` `[1];` `[2], [1,1];` `[3], [1,1,1];` `[4], [2,2], [1,1,1,1];` `[5], [1,1,1,1,1];` `[6], [3,3], [2,2,2], [1,1,1,1,1,1];` `[7], [1,1,1,1,1,1,1];` `[8], [4,4], [2,2,2,2], [1,1,1,1,1,1,1,1];` `[9], [3,3,3], [1,1,1,1,1,1,1,1,1];` `[10], [5,5], [2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1];` `[11], [1,1,1,1,1,1,1,1,1,1,1];` `[12], [6,6], [4,4,4], [3,3,3,3], [2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1];` `[13], [1,1,1,1,1,1,1,1,1,1,1,1,1];` `[14], [7,7], [2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1];` `[15], [5,5,5], [3,3,3,3,3], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];` `[16], [8,8], [4,4,4,4], [2,2,2,2,2,2,2,2], [1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1];` `...` `For n = 6 the 11 partitions of 6 are [6], [3, 3], [4, 2], [2, 2, 2], [5, 1], [3, 2], [4, 1, 1], [2, 2, 1, 1], [3, 1, 1, 1], [2, 1, 1, 1, 1], [1, 1, 1, 1, 1, 1]. There are only four partitions of 6 that contain equal parts so the 6th row of triangle is [6], [3, 3], [2, 2, 2], [1, 1, 1, 1, 1, 1]. The number of parts equals sigma(6) =A000203(6) = 12. The row sum isA038040(6) = 6A000005(6) = 64 = 24.` `FromOmar E. Pol,Dec 04 2019: (Start)` `The structure of the above triangle is as follows:` `1;` `2 11;` `3 111;` `4 22 1111;` `5 11111;` `6 33 222 111111;` `7 1111111;` `8 44 2222 11111111;` `9 333 111111111;` `... (End)`
	MATHEMATICA	`A244051row[n_]:=Flatten[Map[ConstantArray[#, n/#]&, Reverse[Divisors[n]]]];` `Array[A244051row, 10] (Paolo Xausa,Oct 16 2023 )`
	PROG	`(PARI) tabf(nn) = {for (n=1, nn, d = Vecrev(divisors(n)); for (i=1, #d, for (j=1, n/d[i], print1(d[i], "," )); ); print(); ); } \\Michel Marcus,Nov 08 2014`
	CROSSREFS	`Cf.A000005,A000012,A000041,A000203,A000217,A001065,A001227,A002131,A027750,A032742,A038040,A048050,A056538,A126988,A160664,A237593,A245579,A299765,A328365,A309400(mirror).` `Sequence in context:A167269A105535A182980A207974A239454A108888` `Adjacent sequences:A244048A244049A244050A244052A244053A244054`
	KEYWORD	`nonn,tabf,easy`
	AUTHOR	`Omar E. Pol,Nov 08 2014`
	STATUS	`approved`