cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A337737 Least number k such that there are exactly n cubefull numbers between k^3 and (k+1)^3.

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%I A337737 #21 Apr 23 2025 23:14:31
%S A337737 1,2,6,15,12,25,43,73,480,1981,3205,9038,16099,376340,211318,2461230,
%T A337737 2253517,16907618,106308537,312911063
%N A337737 Least number k such that there are exactly n cubefull numbers between k^3 and (k+1)^3.
%C A337737 a(n) = least k such that A337736(k) = n.
%C A337737 Shiu (1991) proved that infinitely many values of k exist for every n. Therefore, this sequence is infinite.
%H A337737 P. Shiu, <a href="https://doi.org/10.1017/S0017089500008351">The distribution of cube-full numbers</a>, Glasgow Mathematical Journal, Vol. 33, No. 3 (1991), pp. 287-295. See section 3, p. 291.
%e A337737 a(0) = 1 since there are no cubefull numbers between 1^3 = 1 and 2^3 = 8.
%e A337737 a(1) = 2 since there is one cubefull number, 16 = 2^4, between 2^3 = 8 and 3^3 = 27.
%e A337737 a(2) = 6 since there are 2 cubefull numbers, 243 = 3^5 and 256 = 2^8, between 6^3 = 216 and 7^3 = 343.
%t A337737 cubQ[n_] := Min[FactorInteger[n][[;; , 2]]] > 2; f[n_] := Count[Range[n^3 + 1, (n + 1)^3 - 1], _?cubQ]; mx = 8; s = Table[0,{mx}]; c = 0; n = 1; While[c < mx, i = f[n] + 1; If[i <= mx && s[[i]] == 0, c++; s[[i]] = n]; n++] ;s
%o A337737 (Python)
%o A337737 from math import gcd
%o A337737 from sympy import integer_nthroot, factorint
%o A337737 def A337737(n):
%o A337737     if n == 0: return 1
%o A337737     a, k = 0, 1
%o A337737     while True:
%o A337737         m, c = k**3, 0
%o A337737         for x in range(1,integer_nthroot(m,5)[0]+1):
%o A337737             if all(d<=1 for d in factorint(x).values()):
%o A337737                 for y in range(1,integer_nthroot(z:=m//x**5,4)[0]+1):
%o A337737                     if gcd(x,y)==1 and all(d<=1 for d in factorint(y).values()):
%o A337737                         c += integer_nthroot(z//y**4,3)[0]
%o A337737         if c-a-1 == n:
%o A337737             return k-1
%o A337737         k += 1
%o A337737         a = c # _Chai Wah Wu_, Apr 23 2025
%Y A337737 Cf. A000578, A036966, A119242, A337736.
%K A337737 nonn,more
%O A337737 0,2
%A A337737 _Amiram Eldar_, Sep 17 2020
%E A337737 a(12)-a(16) from _David A. Corneth_, Sep 18 2020
%E A337737 a(17)-a(19) from _Bert Dobbelaere_, Sep 19 2020