2018
DOI: 10.1007/s12044-018-0418-z
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Universal formulas for the number of partitions

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Cited by 5 publications
(3 citation statements)
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“…in which the series is absolutely convergent. Recently; Aleksa Srdanov (see [19]) investigated the arithmetical function p(n). First he defines numbers p(n, m) the number of all possible partitions of the number n having exactly m parts, (1 m n).…”
Section: Explicit Formula Of Reciprocal Partition Polynomialsmentioning
confidence: 99%
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“…in which the series is absolutely convergent. Recently; Aleksa Srdanov (see [19]) investigated the arithmetical function p(n). First he defines numbers p(n, m) the number of all possible partitions of the number n having exactly m parts, (1 m n).…”
Section: Explicit Formula Of Reciprocal Partition Polynomialsmentioning
confidence: 99%
“…Then p(n) = n k=1 p(n, k). Finally he computed in different way the expression of numbers p(n, k); for more details, we refer to Theorems 1,2,3 and 4 in the work [19]. For S = N, we have…”
Section: Explicit Formula Of Reciprocal Partition Polynomialsmentioning
confidence: 99%
“…, R. STEFANOVIĆ, A. JANKOVIĆ AND D. MILOVANOVIĆof determining the function tenth class of a partition of a natural number n[4].x 1 + x 2 + x 3 + x 4 + x 5 + x 6 + x 7 + x 8 + x 9 + x 10 = 0 2521 9 x 1 + 2521 8 x 2 + 2521 7 x 3 + 2521 6 x 4 + 2521 5 x 5 + 2521 4 x 6 + 2521 3 x 7+ 2521 2 x 8 + 2521x 9 + x 10 = 3323461098288212847 5041 9 x 1 + 5041 8 x 2 + 5041 7 x 3 + 5041 6 x 4 + 5041 5 x 5 + 5041 4 x 6 + 5041 3 x 7 + 5041 2 x 8 + 5041x 9 + x 10 = 1646835119838159208169 7561 9 x 1 + 7561 8 x 2 + 7561 7 x 3 + 7561 6 x 4 + 7561 5 x 5 + 7561 4 x 6 + 7561 3 x 7 + 7561 2 x 8 + 7561x 9 + x 10 = 62620206438528398665452 10081 9 x 1 + 10081 8 x 2 + 10081 7 x 3 + 10081 6 x 4 + 10081 5 x 5 + 10081 4 x 6 + x 7 + 10081 2 x 8 + 10081x 9 + x 10 = 62620206438528398665452 12601 9…”
mentioning
confidence: 99%