A345577 -id:A345577 - cp's OEIS frontend

A003342 Numbers that are the sum of 8 positive 4th powers.

8, 23, 38, 53, 68, 83, 88, 98, 103, 113, 118, 128, 133, 148, 163, 168, 178, 183, 193, 198, 213, 228, 243, 248, 258, 263, 278, 293, 308, 323, 328, 338, 343, 353, 358, 368, 373, 388, 403, 408, 418, 423, 433, 438, 453, 468, 483, 488, 498, 503, 518, 533, 548, 563, 568

Offset: 1

N. J. A. Sloane

			From _David A. Corneth_, Aug 04 2020: (Start)
5396 is in the sequence as 5396 = 1^4 + 1^4 + 4^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4.
8789 is in the sequence as 8789 = 5^4 + 5^4 + 5^4 + 5^4 + 6^4 + 6^4 + 6^4 + 7^4.
12469 is in the sequence as 12469 = 1^4 + 3^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 10^4. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
Eric Weisstein's World of Mathematics, Biquadratic Number.

Cf. A000583, A003331, A003341, A003343, A003353, A345577, A345833.

Select[Range[500], AnyTrue[PowersRepresentations[#, 8, 4], First[#]>0&]&] (* Jean-François Alcover, Jul 18 2017 *)

from itertools import combinations_with_replacement as mc
from sympy import integer_nthroot
def iroot4(n): return integer_nthroot(n, 4)[0]
def aupto(lim):
    pows4 = set(i**4 for i in range(1, iroot4(lim)+1) if i**4 <= lim)
    return sorted(t for t in set(sum(c) for c in mc(pows4, 8)) if t <= lim)
print(aupto(568)) # Michael S. Branicky, Aug 23 2021

A345568 Numbers that are the sum of seven fourth powers in two or more ways.

Original entry on oeis.org

262, 277, 292, 307, 342, 357, 372, 422, 437, 502, 517, 532, 547, 597, 612, 677, 772, 787, 852, 886, 901, 916, 966, 981, 1027, 1046, 1141, 1156, 1221, 1362, 1377, 1396, 1442, 1510, 1525, 1557, 1572, 1587, 1590, 1617, 1637, 1652, 1717, 1765, 1812, 1827, 1892

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			277 is a term because 277 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A003341, A345520, A345559, A345569, A345577, A345605, A345824.

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, 1000)]
for pos in cwr(power_terms, 7):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345578 Numbers that are the sum of eight fourth powers in three or more ways.

Original entry on oeis.org

518, 2678, 2693, 2708, 2738, 2758, 2773, 2838, 2853, 2868, 2883, 2918, 2933, 2948, 2998, 3013, 3078, 3108, 3123, 3173, 3188, 3253, 3302, 3317, 3363, 3382, 3428, 3477, 3492, 3542, 3557, 3622, 3732, 3778, 3797, 3893, 3926, 3953, 3973, 3988, 4018, 4053, 4101

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			2678 is a term because 2678 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 3^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345533, A345569, A345577, A345579, A345587, A345611, A345835.

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, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 3])
    for x in range(len(rets)):
        print(rets[x])

A345586 Numbers that are the sum of nine fourth powers in two or more ways.

Original entry on oeis.org

264, 279, 294, 309, 324, 339, 344, 359, 374, 389, 404, 424, 439, 454, 469, 504, 519, 534, 549, 564, 579, 584, 599, 614, 629, 644, 664, 679, 694, 709, 759, 774, 789, 804, 819, 839, 854, 869, 884, 888, 903, 918, 933, 934, 948, 949, 968, 983, 998, 1013, 1014

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			279 is a term because 279 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A003343, A345541, A345577, A345587, A345595, A345619, A345844.

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, 1000)]
for pos in cwr(power_terms, 9):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

Original entry on oeis.org

263, 278, 293, 308, 323, 343, 358, 373, 388, 423, 438, 453, 503, 533, 548, 563, 583, 598, 613, 628, 678, 693, 758, 773, 788, 803, 853, 868, 887, 902, 917, 932, 933, 967, 982, 997, 1028, 1043, 1047, 1062, 1108, 1127, 1142, 1157, 1172, 1222, 1237, 1283, 1302

Offset: 1

David Consiglio, Jr., Jun 26 2021

nonn

Differs from A345577 at term 14 because 518 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

			278 is a term because 278 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A345577, A345784, A345824, A345833, A345835, A345844, A346327.

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, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 2])
    for x in range(len(rets)):
        print(rets[x])

A345532 Numbers that are the sum of eight cubes in two or more ways.

Original entry on oeis.org

132, 139, 158, 160, 167, 174, 181, 186, 193, 195, 197, 200, 212, 216, 219, 223, 230, 237, 238, 244, 249, 251, 256, 258, 263, 265, 270, 272, 275, 277, 282, 284, 286, 288, 289, 291, 293, 296, 298, 300, 301, 303, 307, 308, 310, 312, 314, 315, 317, 319, 321, 322

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			139 is a term because 139 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 4^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A003331, A345489, A345520, A345533, A345541, A345577, A345784.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**3 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])

A345610 Numbers that are the sum of eight fifth powers in two or more ways.

Original entry on oeis.org

4100, 4131, 4162, 4193, 4342, 4373, 4404, 4584, 4615, 4826, 5123, 5154, 5185, 5365, 5396, 5607, 6146, 6177, 6388, 7169, 7224, 7255, 7286, 7466, 7497, 7708, 8247, 8278, 8489, 9270, 10348, 10379, 10590, 11371, 11875, 11906, 11937, 12117, 12148, 12359, 12898

Offset: 1

David Consiglio, Jr., Jun 20 2021

nonn

			4131 is a term because 4131 = 1^5 + 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 4^5 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5.

Sean A. Irvine, Table of n, a(n) for n = 1..10000

Cf. A003353, A345577, A345605, A345611, A345619, A346327.

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 2])
    for x in range(len(rets)):
        print(rets[x])

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A003342 Numbers that are the sum of 8 positive 4th powers.

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A345568 Numbers that are the sum of seven fourth powers in two or more ways.

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A345578 Numbers that are the sum of eight fourth powers in three or more ways.

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A345586 Numbers that are the sum of nine fourth powers in two or more ways.

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A345834 Numbers that are the sum of eight fourth powers in exactly two ways.

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A345532 Numbers that are the sum of eight cubes in two or more ways.

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A345610 Numbers that are the sum of eight fifth powers in two or more ways.

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