A063883 Number of ways writing n as a sum of different Mersenne prime exponents (terms of A000043).
0, 1, 1, 0, 2, 0, 2, 1, 1, 2, 0, 2, 1, 1, 2, 1, 2, 2, 2, 3, 2, 4, 2, 4, 3, 3, 4, 2, 4, 2, 4, 3, 3, 4, 3, 4, 4, 4, 5, 4, 5, 4, 4, 5, 3, 5, 3, 4, 4, 3, 5, 3, 5, 4, 4, 5, 4, 5, 4, 4, 5, 3, 5, 4, 3, 6, 2, 6, 3, 5, 5, 3, 6, 3, 5, 4, 4, 4, 4, 4, 4, 4, 5, 3, 6, 3, 5, 5, 4, 6, 3, 7, 3, 6, 5, 5, 6, 5, 6, 5, 6, 6, 5, 6, 6
Offset: 1
Examples
n = 50 = 2 + 5 + 7 + 17 + 19 = 2 + 17 + 31 = 19 + 31, so a(50) = 3. The first numbers for which the number of these Mersenne-exponent partitions is k = 0, 1, 2, 3, 4, 5, 6, 7, 8 are 1, 2, 5, 20, 22, 39, 66, 92, 107, respectively.
Links
- T. D. Noe, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Maple
N:= 500: # to get the first N terms G:= mul(1+x^i,i=select(t -> numtheory:-mersenne(t)::integer, [$1..N])): S:= series(G,x,N+1): seq(coeff(S,x,n),n=1..N); # Robert Israel, Sep 22 2016
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Mathematica
exponents[n_] := Reap[For[k = 1, k <= n, k++, If[PrimeQ[2^k-1], Sow[k]]]][[2, 1]]; r[n_] := Module[{ee, x, xx}, ee = exponents[n]; xx = Array[x, Length[ee]]; Reduce[And @@ (0 <= # <= 1 & /@ xx) && xx.ee == n, xx, Integers]]; a[n_] := Which[rn = r[n]; Head[rn] === Or, Length[rn], Head[rn] === And, 1, Head[rn] === Equal, 1, rn === False, 0, True, Print["error ", rn]]; a[1] = 0; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Feb 05 2014 *) With[{e = MersennePrimeExponent[Range[10]]}, Rest@ CoefficientList[Product[1 + x^e[[i]], {i, 1, Length[e]}], x]] (* Amiram Eldar, Dec 23 2024 *)
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PARI
first(lim)=my(M=[2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203, 2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701, 23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433, 1257787, 1398269, 2976221, 3021377, 6972593, 13466917, 20996011, 24036583, 25964951, 30402457, 32582657, 37156667], x='x); if(lim>M[#M], error("Need more Mersenne exponents to compute further")); M=select(p->p<=lim, M); Vec(prod(i=1, #M, 1+x^M[i], O(x^(lim\1+1))+1)) \\ Charles R Greathouse IV, Mar 08 2017
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PARI
a(n) = sum(k=1, 2^n+1, sigma(k)==2^n); \\ Michel Marcus, Mar 07 2017
Formula
a(n) = A054973(2^n). - Michel Marcus, Mar 08 2017
Comments