A093952 -id:A093952 - cp's OEIS frontend

A121919 Least m such that partition number of m modulo m (=A093952(m)) is n.

1, 4, 5, 9, 74, 6, 8, 16, 17, 14, 13, 15, 22, 23, 1402, 19, 41, 69, 26, 232, 61, 617, 28, 38, 30, 205, 50, 196, 65, 32, 175, 56, 96, 381, 45, 140, 57, 104, 59, 51, 119, 795, 262, 117, 78, 88, 86, 60, 106, 812, 113, 63, 81, 90, 229, 72, 66, 209, 71, 68, 352, 178, 64, 354

Offset: 0

Zak Seidov, Sep 02 2006

nonn

			a(3)=9 because partition number of 9 is 30 == 3 modulo 9,
a(5)=74 because partition number of 74 is 7089500 == 5 modulo 74, etc.

Max Alekseyev, Table of n, a(n) for n = 0..10395
Robert G. Wilson v and Max Alekseyev, Table of n, a(n) for n = 0..100000, with unknown terms marked by -1 (contains values below 10^8).

Cf. A000041, A093952.

t = Table[0, {10000}]; k = 1; While[k < 475000, a = Mod[ PartitionsP@k, k]; If[a < 10001 && t[[a]] == 0, t[[a]] = k; Print[{a, k}]]; k++ ]; t (* Robert G. Wilson v, Jul 16 2009 *)

b-file extended by Max Alekseyev, Jun 13 2011, May 19 2014

A051177 Perfectly partitioned numbers: numbers k that divide the number of partitions p(k).

Original entry on oeis.org

1, 2, 3, 124, 158, 342, 693, 1896, 3853, 4434, 5273, 8640, 14850, 17928, 110516, 178984, 274534

Offset: 1

M.A. Muller (mam(AT)land.sun.ac.za)

Are there infinitely many perfectly partitioned numbers? Does there exist some k > 3 for which p(k) is a perfectly partitioned number?
No other terms below 10^8. - Max Alekseyev, May 19 2014
A probabilistic analysis suggests that there are infinitely many terms. - Franklin T. Adams-Watters, Oct 07 2018

			a(4) = 124 because p(124) = 2841940500 is divisible by 124.
a(7) = 693 because partition number of 693 is 43397921522754943172592795 = 693*62623263380598763596815.

Problem 2464, Journal of Recreational Mathematics 29(4), p. 304.
Solution to problem 2464 "Perfect Partitions", Journal of Recreational Mathematics 30(4), pp. 294-295, 1999-2000.

Carlos Rivera, Puzzle 1029. p that divides the number of partitions of p, The Prime Puzzles and Problems Connection.

Cf. A000041.
Cf. A093952 = partition number A000041(n) mod n.
Cf. A056848, A128836, A121015.

Do[ If[ Mod[ PartitionsP@n, n] == 0, Print@n], {n, 250000}] (* Robert G. Wilson v *)
Select[Range[275000],Divisible[PartitionsP[#],#]&] (* Harvey P. Dale, Aug 21 2013~ *)

for(n=1,20000,if(numbpart(n)%n==0,print1(n,","))) \\ Klaus Brockhaus, Sep 06 2006

More terms from Don Reble, Jul 26 2002

A128836 Numbers k such that partition number p(k) == 1 (mod k).

Original entry on oeis.org

1, 4, 7, 11, 54, 55, 115, 146, 157, 234, 239, 951, 272732, 419192, 7626972, 38355152

Offset: 1

Max Alekseyev, Klaus Brockhaus, Robert G. Wilson v and Zak Seidov, Mar 28 2007

Numbers k such that A093952(k) = 1.

There are no other terms below 10^8. - Max Alekseyev, May 19 2014

Eric Weisstein's World of Mathematics, Partition Function P

Cf. A000041, A051177, A093952, A121015, A203023.

a(16) from Max Alekseyev, Jan 15 2013

A094252 a(n) = partition(n) mod prime(n).

Original entry on oeis.org

1, 2, 3, 5, 7, 11, 15, 3, 7, 13, 25, 3, 19, 6, 35, 19, 2, 19, 21, 59, 62, 54, 10, 62, 18, 12, 23, 80, 96, 67, 111, 96, 5, 78, 132, 8, 128, 98, 123, 143, 12, 141, 40, 98, 90, 88, 53, 93, 97, 187, 186, 47, 2, 117, 241, 34, 27, 51, 266, 108, 259, 115, 278, 30, 281, 227, 244, 141

Offset: 1

Gary W. Adamson, Apr 25 2004

nonn

Indices n such that a(n)=0 (i.e., prime(n) divides partition(n)) are listed in A245662. - Max Alekseyev, Jul 27 2014

			a(10) = 13: partition(10) = 42, prime(10) = 29. 42 mod 29 = 13.

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A000041, A093952, A245525.

Table[ Mod[ PartitionsP[n], Prime[n]], {n, 70}] (* Robert G. Wilson v, Apr 28 2004 *)

Edited by Robert G. Wilson v, Apr 28 2004

A121015 Numbers n such that partition number p(n) == 14 (mod n).

Original entry on oeis.org

1, 2, 8, 1402, 3579, 4111, 5289, 6383, 6467, 15146, 32141, 41910, 82849, 110088, 127531, 185114, 1320338, 1467242, 5739729, 22507473, 32494198

Offset: 1

Zak Seidov, Sep 02 2006

No other terms below 10^8. - Max Alekseyev, May 19 2014

			Partition number of 8 is 22 = 1*8 + 14, hence 8 is a term.
Partition number of 1402 is 52435757789401123913939450130086135644 = 37400683159344596229628709079947315*1402 + 14, hence 1402 is a term.

Cf. A000041, A051177, A093952, A128836, A203023.

Do[ If[ Mod[ PartitionsP@n - 14, n] == 0, Print@n], {n, 731000}] (* Robert G. Wilson v, Sep 14 2006 *)

for(n=1,200000,if((numbpart(n)-14)%n==0,print1(n,","))) \\ Klaus Brockhaus, Sep 07 2006

Edited, corrected and extended (a(1) to a(3), a(11) to a(16)) by Klaus Brockhaus, Sep 07 2006
Rechecked by Klaus Brockhaus, Mar 17 2007
a(17)-a(19) from Ryan Propper, Mar 17 2007
a(20) from Max Alekseyev, Dec 28 2011
a(21) from Max Alekseyev, Jan 15 2013

A203023 Integers n dividing A000041(n)+1.

Original entry on oeis.org

1, 6, 156, 305, 484, 1219, 322733, 14343797, 58460571, 68355787

Offset: 1

Max Alekseyev, Dec 27 2011

No other terms below 10^8.

Cf. A093952, A051177, A128836, A121015

A363252 a(n) = gcd(A000041(n), A000009(n)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 5, 2, 2, 2, 4, 1, 1, 1, 1, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 3, 2, 2, 1, 4, 2, 3, 7, 2, 3, 1, 1, 1, 1, 21, 21, 2, 1, 1, 2, 6, 14, 3, 1, 2, 1, 1, 1, 1, 2, 1, 3, 4, 4, 17, 1, 2, 1, 2, 2, 4, 1, 3, 5, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 4, 1, 1, 1, 2, 11, 2

Offset: 0

Vaclav Kotesovec, May 23 2023, inspired by Zhi-Wei Sun

nonn

Cf. A000009, A000041, A051177, A056848, A093952, A304991.

b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*add(
     `if`(d::odd, d, 0), d=numtheory[divisors](j)), j=1..n)/n)
    end:
a:= n-> igcd(b(n), combinat[numbpart](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 23 2023

Table[GCD[PartitionsP[n], PartitionsQ[n]], {n, 0, 100}]

cp's OEIS Frontend

A121919 Least m such that partition number of m modulo m (=A093952(m)) is n.

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A051177 Perfectly partitioned numbers: numbers k that divide the number of partitions p(k).

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A128836 Numbers k such that partition number p(k) == 1 (mod k).

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A094252 a(n) = partition(n) mod prime(n).

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A121015 Numbers n such that partition number p(n) == 14 (mod n).

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A203023 Integers n dividing A000041(n)+1.

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A363252 a(n) = gcd(A000041(n), A000009(n)).

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