A257989 The crank of the partition having Heinz number n.
-1, 2, -2, 3, 0, 4, -3, 2, 0, 5, -2, 6, 0, 3, -4, 7, 1, 8, -1, 4, 0, 9, -3, 3, 0, 2, -1, 10, 1, 11, -5, 5, 0, 4, -2, 12, 0, 6, -3, 13, 1, 14, -1, 3, 0, 15, -4, 4, 1, 7, -1, 16, 2, 5, -2, 8, 0, 17, -1, 18, 0, 4, -6, 6, 1, 19, -1, 9, 1, 20, -3, 21, 0, 3, -1, 5, 1, 22, -4, 2, 0, 23, -1, 7, 0, 10, -2, 24, 2, 6, -1
Offset: 2
Keywords
Examples
a(12) = - 2 because the partition with Heinz number 12 = 2*2*3 is [1,1,2], the number of parts larger than the number of 1's is 0 and the number of 1's is 2; 0 - 2 = -2. a(945) = 4 because the partition with Heinz number 945 = 3^3 * 5 * 7 is [2,2,2,3,4] which has no part 1; the largest part is 4. From _Gus Wiseman_, Apr 05 2021: (Start) The partitions (center) with each Heinz number (left), and the corresponding terms (right): 2: (1) -> -1 3: (2) -> 2 4: (1,1) -> -2 5: (3) -> 3 6: (2,1) -> 0 7: (4) -> 4 8: (1,1,1) -> -3 9: (2,2) -> 2 10: (3,1) -> 0 11: (5) -> 5 12: (2,1,1) -> -2 13: (6) -> 6 14: (4,1) -> 0 15: (3,2) -> 3 16: (1,1,1,1) -> -4 (End)
Links
- Alois P. Heinz, Table of n, a(n) for n = 2..10000
- G. E. Andrews and F. Garvan, Dyson's crank of a partition, Bull. Amer. Math. Soc., 18 (1988), 167-171.
- B. C. Berndt, H. H. Chan, S. H. Chan, W.-C. Liaw, Cranks and dissections in Ramanujan's lost notebook, J. Comb. Theory, Ser. A, 109, 2005, 91-120.
- B. C. Berndt, H. H. Chan, S. H. Chan, W.-C. Liaw, Cranks - really the final problem, Ramanujan J., 23, 2010, 3-15.
- G. E. Andrews, K. Ono, Ramanujan's congruences and Dyson's crank, Proc. Natl. Acad. Sci. USA, 102, 2005, 15277.
- FindStat, Dyson's crank of a partition.
- K. Mahlburg, Partition congruences and the Andrews-Garvan-Dyson crank, Proc. Natl. Acad. Sci. USA, 102, 2005, 15373-15376.
- Wikipedia, Crank of a partition
Crossrefs
Programs
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Maple
with(numtheory): a := proc (n) local B, b, c: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do; [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: b := proc (n) if `mod`(n, 2) = 1 then 0 else 1+b((1/2)*n) end if end proc: c := proc (n) local b, B, ct, i: b := proc (n) if `mod`(n, 2) = 1 then 0 else 1+b((1/2)*n) end if end proc: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: ct := 0: for i to bigomega(n) do if b(n) < B(n)[i] then ct := ct+1 else end if end do: ct end proc: if b(n) = 0 then max(B(n)) else c(n)-b(n) end if end proc: seq(a(n), n = 2 .. 150);
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Mathematica
B[n_] := Module[{nn, j, m}, nn = FactorInteger[n]; For[j = 1, j <= Length[nn], j++, m[j] = nn[[j]]]; Flatten[Table[Table[PrimePi[m[i][[1]]], {q, 1, m[i][[2]]}], {i, 1, Length[nn]}]]]; b[n_] := b[n] = If[OddQ[n], 0, 1 + b[n/2]]; c[n_] := Module[{ct, i}, ct = 0; For[i = 1, i <= PrimeOmega[n], i++, If[ b[n] < B[n][[i]], ct++]]; ct]; a[n_] := If[b[n] == 0, Max[B[n]], c[n] - b[n]]; Table[a[n], {n, 2, 100}] (* Jean-François Alcover, Apr 25 2017, after Emeric Deutsch *) primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; ck[y_]:=With[{w=Count[y,1]},If[w==0,Max@@y,Count[y,_?(#>w&)]-w]]; Table[ck[primeMS[n]],{n,2,30}] (* Gus Wiseman, Apr 05 2021 *)
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