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 1-6 of 6 results.

A345563 Numbers that are the sum of six fourth powers in six or more ways.

Original entry on oeis.org

21251, 37811, 38051, 43251, 43571, 43875, 44115, 44531, 45155, 45651, 45891, 47411, 47586, 48276, 49796, 49971, 52195, 53235, 53315, 54131, 56290, 57395, 57460, 57570, 58035, 58500, 59075, 59330, 59780, 59795, 59811, 59860, 60035, 62180, 62211, 63971, 66340
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			37811 is a term because 37811 = 1^4 + 2^4 + 2^4 + 7^4 + 11^4 + 12^4 = 2^4 + 2^4 + 4^4 + 7^4 + 9^4 + 13^4 = 2^4 + 3^4 + 6^4 + 6^4 + 9^4 + 13^4 = 3^4 + 4^4 + 8^4 + 8^4 + 11^4 + 11^4 = 4^4 + 6^4 + 7^4 + 9^4 + 9^4 + 12^4 = 5^4 + 5^4 + 9^4 + 10^4 + 10^4 + 10^4.
		

Crossrefs

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, 1000)]
    for pos in cwr(power_terms, 6):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 6])
        for x in range(len(rets)):
            print(rets[x])

A345609 Numbers that are the sum of seven fifth powers in six or more ways.

Original entry on oeis.org

13562501, 14583968, 21555313, 22057487, 22066065, 23089782, 23345024, 24217918, 24401574, 24855016, 24952718, 24993517, 25052501, 25385064, 28608832, 29558618, 30653536, 31613713, 32559143, 33005785, 33533765, 33635825, 33828631, 34267551, 34268332, 35431351
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			14583968 is a term because 14583968 = 1^5 + 4^5 + 14^5 + 16^5 + 19^5 + 21^5 + 23^5 = 2^5 + 4^5 + 14^5 + 14^5 + 20^5 + 22^5 + 22^5 = 4^5 + 5^5 + 10^5 + 15^5 + 20^5 + 21^5 + 23^5 = 6^5 + 8^5 + 9^5 + 15^5 + 15^5 + 20^5 + 25^5 = 6^5 + 8^5 + 14^5 + 14^5 + 14^5 + 16^5 + 26^5 = 6^5 + 10^5 + 12^5 + 12^5 + 16^5 + 16^5 + 26^5.
		

Crossrefs

Programs

  • Python
    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, 7):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 6])
        for x in range(len(rets)):
            print(rets[x])

A345719 Numbers that are the sum of six fifth powers in five or more ways.

Original entry on oeis.org

54827300, 74115800, 74883600, 75609125, 113088250, 120274275, 166078869, 169692136, 174781858, 178736448, 182341225, 185558208, 194939538, 203054589, 218814275, 235067008, 250989825, 251772882, 252721458, 255453233, 258124975, 274616694, 282859667
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			74115800 is a term because 74115800 = 1^5 + 4^5 + 21^5 + 21^5 + 29^5 + 34^5 = 1^5 + 8^5 + 14^5 + 23^5 + 32^5 + 32^5 = 4^5 + 11^5 + 13^5 + 22^5 + 24^5 + 36^5 = 5^5 + 6^5 + 19^5 + 20^5 + 24^5 + 36^5 = 6^5 + 25^5 + 25^5 + 25^5 + 29^5 + 30^5.
		

Crossrefs

Programs

  • Python
    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, 6):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 5])
        for x in range(len(rets)):
            print(rets[x])

A345721 Numbers that are the sum of six fifth powers in seven or more ways.

Original entry on oeis.org

1184966816, 1700336000, 1717860100, 1972000800, 2229475325, 2295937600, 2396275200, 2548597632, 2625460992, 2886251808, 3217068800, 3697267200, 3729261536, 3765398725, 4046532448, 4165116967, 4246566632, 4286704224, 4335900525, 4489548050
Offset: 1

Views

Author

David Consiglio, Jr., Jun 24 2021

Keywords

Examples

			1700336000 is a term because 1700336000 = 4^5 + 17^5 + 31^5 + 37^5 + 43^5 + 68^5 = 6^5 + 9^5 + 10^5 + 23^5 + 60^5 + 62^5 = 6^5 + 14^5 + 16^5 + 50^5 + 50^5 + 64^5 = 7^5 + 25^5 + 30^5 + 54^5 + 56^5 + 58^5 = 8^5 + 21^5 + 23^5 + 27^5 + 57^5 + 64^5 = 9^5 + 21^5 + 22^5 + 29^5 + 53^5 + 66^5 = 13^5 + 32^5 + 35^5 + 38^5 + 45^5 + 67^5.
		

Crossrefs

Programs

  • Python
    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, 6):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 7])
        for x in range(len(rets)):
            print(rets[x])

A345864 Numbers that are the sum of five fifth powers in six or more ways.

Original entry on oeis.org

288203194368, 2784485221024, 6022068333568, 9222502219776, 9670153077344, 10918228000032, 15441787364576
Offset: 1

Views

Author

David Consiglio, Jr., Jun 27 2021

Keywords

Crossrefs

Programs

  • Python
    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, 5):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 6])
        for x in range(len(rets)):
            print(rets[x])

A346361 Numbers that are the sum of six fifth powers in exactly six ways.

Original entry on oeis.org

287718651, 553545456, 746783675, 972232800, 1005620508, 1040741042, 1070652352, 1074892544, 1182426366, 1197332400, 1243267146, 1317183650, 1364866263, 1387455091, 1429663400, 1498160992, 1529189818, 1554833117, 1558594400, 1610298901, 1623782765, 1627228231
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345720 at term 10 because 1184966816 = 15^5 + 24^5 + 27^5 + 38^5 + 39^5 + 63^5 = 2^5 + 28^5 + 36^5 + 36^5 + 42^5 + 62^5 = 4^5 + 24^5 + 38^5 + 38^5 + 40^5 + 62^5 = 21^5 + 32^5 + 37^5 + 41^5 + 45^5 + 60^5 = 8^5 + 14^5 + 34^5 + 40^5 + 52^5 + 58^5 = 11^5 + 17^5 + 22^5 + 49^5 + 51^5 + 56^5 = 11^5 + 16^5 + 22^5 + 52^5 + 52^5 + 53^5.

Examples

			287718651 is a term because 287718651 = 10^5 + 11^5 + 20^5 + 22^5 + 30^5 + 48^5 = 8^5 + 10^5 + 21^5 + 27^5 + 27^5 + 48^5 = 3^5 + 6^5 + 25^5 + 30^5 + 30^5 + 47^5 = 9^5 + 10^5 + 13^5 + 26^5 + 37^5 + 46^5 = 6^5 + 9^5 + 14^5 + 31^5 + 35^5 + 46^5 = 10^5 + 11^5 + 12^5 + 23^5 + 41^5 + 44^5.
		

Crossrefs

Programs

  • Python
    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, 6):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 6])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-6 of 6 results.