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.

Showing 1-5 of 5 results.

A344188 Numbers that are the sum of three fourth powers in exactly one way.

Original entry on oeis.org

3, 18, 33, 48, 83, 98, 113, 163, 178, 243, 258, 273, 288, 338, 353, 418, 513, 528, 593, 627, 642, 657, 707, 722, 768, 787, 882, 897, 962, 1137, 1251, 1266, 1298, 1313, 1328, 1331, 1378, 1393, 1458, 1506, 1553, 1568, 1633, 1808, 1875, 1922, 1937, 2002, 2177, 2403, 2418, 2433, 2483, 2498, 2546, 2563, 2593, 2608, 2658
Offset: 1

Views

Author

David Consiglio, Jr., May 11 2021

Keywords

Comments

Differs from A003337 and A047714 at term 60 because 2673 = 2^4 + 4^4 + 7^4 = 3^4 + 6^4 + 6^4, see A309762.

Examples

			33 is a member of this sequence because 33 = 1^4 + 2^4 + 2^4

Links

Crossrefs

Cf. A003337, A025395, A344187, A344189, A344192, A344641.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**4 for x in range(1,50)]
for pos in cwr(power_terms,3):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v == 1])
for x in range(len(rets)):
    print(rets[x])

A047714 Sums of 3 but no fewer nonzero fourth powers.

Original entry on oeis.org

3, 18, 33, 48, 83, 98, 113, 163, 178, 243, 258, 273, 288, 338, 353, 418, 513, 528, 593, 627, 642, 657, 707, 722, 768, 787, 882, 897, 962, 1137, 1251, 1266, 1298, 1313, 1328, 1331, 1378, 1393, 1458, 1506, 1553, 1568, 1633, 1808, 1875, 1922, 1937, 2002, 2177
Offset: 1

Views

Author

Arlin Anderson (starship1(AT)gmail.com)

Keywords

Comments

Identical to A003337 for n = 1..87. - Michael S. Branicky, Mar 18 2021

Links

Crossrefs

Cf. A000583, A002377, A336536, A344187.
Subsequence of A003337.

Programs

  • Maple
    N:= 3000: # for terms <= N
F1:= {seq(i^4,i=1..floor(N^(1/4)))}: n1:= nops(F1):
F2:= select(`<=`,{seq(seq(F1[i]+F1[j],i=1..j),j=1..nops(F1))},N):
F3:= select(`<=`,{seq(seq(s+t,s=F1),t=F2)},N):
A:= sort(convert(F3 minus (F2 union F1), list)); # Robert Israel, Jul 24 2020
  • Python
    def aupto(lim):
  p1 = set(i**4 for i in range(1, int(lim**.25)+2) if i**4 <= lim)
  p2 = set(a+b for a in p1 for b in p1 if a+b <= lim)
  p3 = set(apb+c for apb in p2 for c in p1 if apb+c <= lim)
  return sorted(p3-p2-p1)
print(aupto(2400)) # Michael S. Branicky, Mar 18 2021

Formula

A002377(a(n)) = 3. - Robert Israel, Jul 24 2020
Equals A003337 - A344187 - A000583. - Sean A. Irvine, May 15 2021

A338667 Numbers that are the sum of two positive cubes in exactly one way.

Original entry on oeis.org

2, 9, 16, 28, 35, 54, 65, 72, 91, 126, 128, 133, 152, 189, 217, 224, 243, 250, 280, 341, 344, 351, 370, 407, 432, 468, 513, 520, 539, 559, 576, 637, 686, 728, 730, 737, 756, 793, 854, 855, 945, 1001, 1008, 1024, 1027, 1064, 1072, 1125, 1216, 1241, 1332, 1339, 1343, 1358, 1395, 1456, 1458, 1512, 1547, 1674, 1736
Offset: 1

Views

Author

David Consiglio, Jr., Apr 22 2021

Keywords

Comments

This sequence differs from A003325 at term 61: A003325(61) = 1729 is the famous Ramanujan taxicab number and is excluded from this sequence because it is the sum of two cubes in two distinct ways.

Examples

			35 is a term of this sequence because 2^3 + 3^3 = 8 + 27 = 35 and this is the one and only way to express 35 as the sum of two cubes.

Links

Crossrefs

Cf. A003325, A025284, A025395, A343708, A344187.

Programs

  • Mathematica
    Select[Range@2000,Length[s=PowersRepresentations[#,2,3]]==1&&And@@(#>0&@@@s)&] (* Giorgos Kalogeropoulos, Apr 24 2021 *)
  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
from bisect import bisect_left as bisect
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1,1000)]
for pos in cwr(power_terms,2):
    tot = sum(pos)
    keep[tot] += 1
rets = sorted([k for k,v in keep.items() if v == 1])
for x in range(len(rets)):
    print(rets[x])

A047716 Sums of 5 but no fewer nonzero fourth powers.

Original entry on oeis.org

5, 20, 35, 50, 65, 80, 85, 100, 115, 130, 145, 165, 180, 195, 210, 245, 260, 275, 290, 305, 320, 325, 340, 355, 370, 385, 405, 420, 435, 450, 500, 515, 530, 545, 560, 580, 595, 610, 629, 644, 659, 675, 689, 690, 709, 724, 739, 754, 755, 770, 785, 789, 800
Offset: 1

Views

Author

Arlin Anderson (starship1(AT)gmail.com)

Keywords

Links

Crossrefs

Subsequence of A003339.
Cf. A000583, A002377, A344187, A344188, A344189.

Programs

  • PARI
    upto(n)={my(e=5); my(s=sum(k=1, sqrtint(sqrtint(n)), x^(k^4)) + O(x*x^n)); my(p=s^e, q=(1 + s)^(e-1)); select(k->polcoeff(p,k) && !polcoeff(q,k), [1..n])} \\ Andrew Howroyd, Jul 06 2018
  • Python
    from itertools import combinations_with_replacement as combs_with_rep
def aupto(limit):
  qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 4 <= limit]
  ss = [set(sum(c) for c in combs_with_rep(qd, k)) for k in range(6)]
  return sorted(filter(lambda x: x <= limit, ss[5]-ss[4]-ss[3]-ss[2]-ss[1]))
print(aupto(800)) # Michael S. Branicky, May 20 2021

Formula

Equals A003339 - A344189 - A344188 - A344187 - A000583. - Sean A. Irvine, May 15 2021

A047715 Numbers that are the sum of 4 but no fewer nonzero fourth powers.

Original entry on oeis.org

4, 19, 34, 49, 64, 84, 99, 114, 129, 164, 179, 194, 244, 259, 274, 289, 304, 324, 339, 354, 369, 419, 434, 499, 514, 529, 544, 594, 609, 628, 643, 658, 673, 674, 708, 723, 738, 769, 784, 788, 803, 849, 868, 883, 898, 913, 963, 978, 1024, 1043, 1138, 1153
Offset: 1

Views

Author

Arlin Anderson (starship1(AT)gmail.com)

Keywords

Comments

First differs from A003338 at term 64: A003338(64) = 1393 is also a term of A003337, so not a term here. - Michael S. Branicky, Apr 19 2021

Crossrefs

Cf. A000583, A002377, A003338 (sum of 4), A003337 (sum of 3), A003336 (sum of 2), A344188, A344187.

Programs

  • Python
    limit = 1153
from functools import lru_cache
qd = [k**4 for k in range(1, int(limit**.25)+2) if k**4 + 3 <= limit]
qds = set(qd)
@lru_cache(maxsize=None)
def findsums(n, m):
  if m == 1: return {(n,)} if n in qds else set()
  return set(tuple(sorted(t+(q,))) for q in qds for t in findsums(n-q, m-1))
A003338s = set(n for n in range(4, limit+1) if len(findsums(n, 4)) >= 1)
A003337s = set(n for n in range(3, limit+1) if len(findsums(n, 3)) >= 1)
A003336s = set(n for n in range(2, limit+1) if len(findsums(n, 2)) >= 1)
print(sorted(A003338s - A003337s - A003336s - qds)) # Michael S. Branicky, Apr 19 2021

Formula

Equals A003338 - A344188 - A344187 - A000583, where "-" denotes "set difference". - Sean A. Irvine, May 15 2021
Showing 1-5 of 5 results.

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