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 1-4 of 4 results.

A345603 Numbers that are the sum of ten fourth powers in ten or more ways.

Original entry on oeis.org

6885, 7990, 8035, 8100, 8165, 8275, 8340, 8515, 8565, 8580, 9140, 9205, 9235, 9285, 9300, 9315, 9380, 9445, 9495, 9510, 9540, 9555, 9620, 9670, 9685, 9750, 9795, 9830, 9860, 9924, 9925, 9990, 10005, 10164, 10294, 10340, 10374, 10404, 10420, 10515, 10534
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			7990 is a term because 7990 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 6^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 6^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 1^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 9^4 = 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 7^4.
		

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, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 10])
        for x in range(len(rets)):
            print(rets[x])

A345627 Numbers that are the sum of nine fifth powers in ten or more ways.

Original entry on oeis.org

4157156, 4492410, 4510461, 4915538, 4948274, 5005474, 5015506, 5179747, 5219655, 5252477, 5739988, 5756794, 6323426, 6326519, 6382443, 6423394, 6654999, 6705284, 6793170, 6861218, 7101038, 7147645, 7147656, 7148679, 7266240, 7280391, 7283268, 7314187, 7413493
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			4492410 is a term because 4492410 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 7^5 + 13^5 + 21^5 = 1^5 + 2^5 + 6^5 + 10^5 + 11^5 + 11^5 + 14^5 + 16^5 + 19^5 = 1^5 + 6^5 + 7^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 7^5 + 15^5 + 15^5 + 16^5 + 18^5 = 2^5 + 5^5 + 6^5 + 10^5 + 10^5 + 11^5 + 11^5 + 15^5 + 20^5 = 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 + 18^5 = 3^5 + 3^5 + 8^5 + 8^5 + 8^5 + 12^5 + 12^5 + 17^5 + 19^5 = 3^5 + 4^5 + 6^5 + 7^5 + 8^5 + 13^5 + 14^5 + 16^5 + 19^5 = 4^5 + 4^5 + 4^5 + 7^5 + 11^5 + 11^5 + 13^5 + 18^5 + 18^5 = 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 16^5 + 17^5 + 18^5.
		

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, 9):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 10])
        for x in range(len(rets)):
            print(rets[x])

A345641 Numbers that are the sum of ten fifth powers in nine or more ways.

Original entry on oeis.org

1192180, 1226654, 1242437, 1431399, 1431430, 1431641, 1431672, 1431883, 1432453, 1432664, 1434765, 1439174, 1439416, 1441695, 1442718, 1447602, 1448447, 1455346, 1455377, 1464166, 1464377, 1464408, 1474431, 1474462, 1475485, 1491978, 1497619, 1505660, 1531398
Offset: 1

Views

Author

David Consiglio, Jr., Jun 20 2021

Keywords

Examples

			1226654 is a term because 1226654 = 1^5 + 1^5 + 4^5 + 5^5 + 7^5 + 8^5 + 9^5 + 13^5 + 13^5 + 13^5 = 1^5 + 1^5 + 5^5 + 7^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 = 1^5 + 2^5 + 2^5 + 4^5 + 6^5 + 10^5 + 12^5 + 12^5 + 12^5 + 13^5 = 1^5 + 2^5 + 2^5 + 4^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 14^5 = 1^5 + 3^5 + 3^5 + 3^5 + 7^5 + 10^5 + 10^5 + 10^5 + 13^5 + 14^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 7^5 + 8^5 + 12^5 + 13^5 + 14^5 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 5^5 + 10^5 + 13^5 + 13^5 + 13^5 = 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 8^5 + 9^5 + 16^5 = 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 8^5 + 9^5 + 16^5.
		

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, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v >= 9])
        for x in range(len(rets)):
            print(rets[x])

A346355 Numbers that are the sum of ten fifth powers in exactly ten ways.

Original entry on oeis.org

1431641, 1439416, 1464377, 1464408, 1505660, 1531640, 1564165, 1782171, 1969253, 1976997, 1986028, 2000966, 2028270, 2042460, 2052415, 2058421, 2059202, 2060522, 2076393, 2130272, 2201247, 2208681, 2209704, 2248941, 2250329, 2251042, 2282073, 2307747, 2315379
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

Differs from A345642 at term 6 because 1531398 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 10^5 + 12^5 + 16^5 = 1^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 10^5 + 11^5 + 11^5 + 16^5 = 1^5 + 1^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 10^5 + 14^5 + 15^5 = 2^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 10^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 4^5 + 10^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 13^5 + 14^5 + 14^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 13^5 + 14^5 + 14^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5 + 13^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 7^5 + 7^5 + 10^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 6^5 + 7^5 + 10^5 + 12^5 + 12^5 + 13^5 + 14^5.

Examples

			1431641 is a term because 1431641 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 6^5 + 7^5 + 10^5 + 12^5 + 16^5 = 1^5 + 1^5 + 4^5 + 6^5 + 7^5 + 7^5 + 8^5 + 9^5 + 12^5 + 16^5 = 1^5 + 3^5 + 3^5 + 5^5 + 6^5 + 7^5 + 8^5 + 11^5 + 11^5 + 16^5 = 1^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 + 15^5 = 1^5 + 1^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 14^5 + 15^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 + 15^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 10^5 + 10^5 + 10^5 + 13^5 + 15^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 + 15^5 = 1^5 + 1^5 + 2^5 + 3^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 + 14^5 = 1^5 + 1^5 + 2^5 + 3^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5 + 14^5.
		

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, 10):
        tot = sum(pos)
        keep[tot] += 1
        rets = sorted([k for k, v in keep.items() if v == 10])
        for x in range(len(rets)):
            print(rets[x])
Showing 1-4 of 4 results.