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.

User: David Consiglio, Jr.

David Consiglio, Jr.'s wiki page.

David Consiglio, Jr. has authored 459 sequences. Here are the ten most recent ones:

A369715 Number of digits of phi (the golden ratio) correctly approximated by Fibonacci(n+1) / Fibonacci(n).

Original entry on oeis.org

1, 0, 1, 2, 2, 2, 3, 3, 3, 4, 3, 5, 5, 6, 6, 6, 7, 6, 8, 8, 9, 8, 10, 10, 10, 11, 11, 11, 12, 11, 13, 13, 14, 13, 14, 15, 16, 15, 16, 17, 17, 17, 18, 18, 18, 19, 19, 20, 19, 21, 21, 22, 22, 22, 23, 23, 24, 24, 25, 24, 25, 26, 26, 27, 27, 28, 28, 28, 29, 29, 30, 30, 30
Offset: 1

Views

Author

David Consiglio, Jr., Jan 31 2024

Keywords

Examples

			For n=1, 1/1 = 1 matches the first digit of phi (1.618033), so a(1) = 1
For n=2, 2/1 = 2 which matches no digits of phi (1.618033), so a(2) = 0
For n=12,
  F(13)/F(12) = 1.6180 55... = 233/144
  phi         = 1.6180 33...
                ^ ^^^^    a(12) = 5 matching digits

Links

Crossrefs

Cf. A000045, A001622, A048433, A048434.

Programs

  • Python
    from math import isqrt
fib = [1,1]
terms = []
while len(terms) < 1000:
    deg = 0
    target = 0
    test = 0
    while target == test:
        target = (10**deg+isqrt(5*10**(2*deg)))//2
        test = (10**deg*(fib[-1]))//fib[-2]
        deg += 1
    terms.append(deg-1)
    fib.append(fib[-1]+fib[-2])
print(terms)

A349031 The "Fouriest" numbers: numbers that can be expressed as a string of 4's longer than the original number in some base.

Original entry on oeis.org

624, 3124, 6220, 15624, 37324, 78124, 78432, 223948, 390624, 549028, 1343692, 1953124, 3843200, 8062156, 9586980, 9765624, 26902404, 48372940, 48828124, 76695844, 188316832, 244140624, 290237644, 613566756, 1220703124, 1318217828, 1741425868, 4908534052
Offset: 1

Views

Author

David Consiglio, Jr., Nov 06 2021

Keywords

Comments

Inspired by Saturday Morning Breakfast Cereal comics.
The "Fouriest" numbers. The number shown in the comment, 624, is correctly identified as a term of the sequence.

Examples

			624 is a member of this sequence because 624 expressed in base 5 is 4444. 4444 has 4 digits and 624 has only 3.

Links

Crossrefs

For the Fouriest transform see A268236-A268238, A268300. - N. J. A. Sloane, Nov 19 2021

Programs

  • Python
    from math import log
def A349031(limit):
    super_fours = []
    for length in range(4,int(log(limit,5))+1):
        fours = "4"*length
        for base in range(5, 10):
            keep = 4*(1-base**length)//(1-base)
            if len(str(keep)) < len(fours) and keep < limit:
                super_fours.append(keep)
    return sorted(super_fours)
result = A349031(10**20)

A346803 Numbers that are the sum of nine squares in ten or more ways.

Original entry on oeis.org

63, 65, 68, 71, 72, 74, 75, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			65 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 6^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 4^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 65 is a term.

Links

Crossrefs

Cf. A025428, A025429, A345497, A345549, A346808.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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])
  • Python
    def A346803(n): return (63, 65, 68, 71, 72, 74, 75)[n-1] if n<8 else n+69 # Chai Wah Wu, May 09 2024

Formula

From Chai Wah Wu, May 09 2024: (Start)
All integers >= 77 are terms. Proof: since 246 can be written as the sum of 4 positive squares in 10 ways (see A025428) and any integer >= 34 can be written as a sum of 5 positive squares (see A025429), any integer >= 280 can be written as a sum of 9 positive squares in 10 or more ways. Integers from 77 to 279 are terms by inspection.
a(n) = 2*a(n-1) - a(n-2) for n > 9.
G.f.: x*(-x^8 + x^7 - x^6 + x^5 - 2*x^4 + x^2 - 61*x + 63)/(x - 1)^2. (End)

A346807 Numbers that are the sum of ten squares in eight or more ways.

Original entry on oeis.org

58, 61, 64, 66, 67, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			61 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 61 is a term.

Links

Crossrefs

Cf. A025434, A345505, A345556, A346806.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 8])
    for x in range(len(rets)):
        print(rets[x])

Formula

a(n) = n + 64 for n >= 8 (conjectured). - Chai Wah Wu, Dec 05 2023

A346806 Numbers that are the sum of ten squares in seven or more ways.

Original entry on oeis.org

57, 58, 60, 61, 63, 64, 66, 67, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			58 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 58 is a term.

Links

Crossrefs

Cf. A345504, A345555, A346805, A346807.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 7])
    for x in range(len(rets)):
        print(rets[x])

A346804 Numbers that are the sum of ten squares in five or more ways.

Original entry on oeis.org

40, 45, 48, 49, 51, 52, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			45 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2
so 45 is a term.

Links

Crossrefs

Cf. A345502, A345553, A346805.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 5])
    for x in range(len(rets)):
        print(rets[x])

A346805 Numbers that are the sum of ten squares in six or more ways.

Original entry on oeis.org

49, 52, 55, 57, 58, 60, 61, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			52 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2
so 52 is a term.

Links

Crossrefs

Cf. A345503, A345554, A346804, A346806.

Programs

  • Mathematica
    Select[Range[120],Count[IntegerPartitions[#,{10}],?(AllTrue[Sqrt[#],IntegerQ]&)]>5&] (* _Harvey P. Dale, Sep 10 2023 *)
  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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 >= 6])
    for x in range(len(rets)):
        print(rets[x])

A346808 Numbers that are the sum of ten squares in ten or more ways.

Original entry on oeis.org

61, 64, 66, 67, 69, 70, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124
Offset: 1

Views

Author

David Consiglio, Jr., Aug 04 2021

Keywords

Examples

			64 = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 7^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 3^2 + 3^2 + 6^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 5^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 4^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 5^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2 + 4^2
   = 1^2 + 1^2 + 1^2 + 1^2 + 1^2 + 3^2 + 3^2 + 3^2 + 4^2 + 4^2
   = 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 4^2 + 4^2
   = 1^2 + 2^2 + 2^2 + 2^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 4^2
   = 1^2 + 1^2 + 2^2 + 2^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2 + 3^2
so 64 is a term.

Links

Crossrefs

Cf. A345558, A346803. Subsequence of A346807.

Programs

  • Python
    from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**2 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])

A346365 Numbers that are the sum of six fifth powers in exactly ten ways.

Original entry on oeis.org

55302546200, 89999127392, 96110537743, 104484239200, 120492759200, 121258798144, 127794946400, 133364991375, 135030535200, 136156575744, 151305014432, 155434423925, 174388570400, 177099008000, 179272687000, 182844944832, 184948721056, 187873845500
Offset: 1

Views

Author

David Consiglio, Jr., Jul 18 2021

Keywords

Comments

This sequence differs from A344196:
180336745600 = 48^5 + 54^5 + 66^5 + 66^5 + 112^5 + 174^5
= 9^5 + 21^5 + 93^5 + 112^5 + 117^5 + 168^5
= 11^5 + 44^5 + 73^5 + 92^5 + 133^5 + 167^5
= 15^5 + 81^5 + 94^5 + 95^5 + 129^5 + 166^5
= 1^5 + 49^5 + 62^5 + 107^5 + 138^5 + 163^5
= 35^5 + 69^5 + 75^5 + 98^5 + 141^5 + 162^5
= 18^5 + 81^5 + 105^5 + 112^5 + 135^5 + 159^5
= 14^5 + 50^5 + 62^5 + 86^5 + 150^5 + 158^5
= 2^5 + 52^5 + 54^5 + 108^5 + 146^5 + 158^5
= 14^5 + 22^5 + 66^5 + 118^5 + 142^5 + 158^5
= 4^5 + 50^5 + 58^5 + 102^5 + 150^5 + 156^5,
so 180336745600 is in A344196, but is not in this sequence.

Examples

			55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5
            = 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5
            = 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5
            = 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5
            = 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5
            = 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5
            = 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5
            = 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5
            = 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5
            = 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5
so 55302546200 is a term.

Links

Crossrefs

Cf. A344196, A345822, A346259, A346364.

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 == 10])
    for x in range(len(rets)):
        print(rets[x])

A346364 Numbers that are the sum of six fifth powers in exactly nine ways.

Original entry on oeis.org

9085584992, 16933805856, 37377003050, 39254220544, 41066625600, 41485873792, 42149876800, 43828403850, 44180505600, 45902654525, 48588434400, 52005184992, 53536896864, 54156285568, 56229189632, 57088402525, 59954496800, 63432407850, 66188522400, 66507304800
Offset: 1

Views

Author

David Consiglio, Jr., Jul 13 2021

Keywords

Comments

This sequence differs from A345723:
55302546200 = 34^5 + 38^5 + 50^5 + 57^5 + 95^5 + 136^5
= 23^5 + 49^5 + 61^5 + 69^5 + 107^5 + 131^5
= 24^5 + 37^5 + 63^5 + 81^5 + 104^5 + 131^5
= 21^5 + 35^5 + 60^5 + 94^5 + 100^5 + 130^5
= 57^5 + 60^5 + 71^5 + 75^5 + 109^5 + 128^5
= 19^5 + 37^5 + 56^5 + 96^5 + 104^5 + 128^5
= 35^5 + 41^5 + 53^5 + 69^5 + 115^5 + 127^5
= 16^5 + 49^5 + 53^5 + 83^5 + 112^5 + 127^5
= 35^5 + 37^5 + 40^5 + 88^5 + 119^5 + 121^5
= 11^5 + 24^5 + 71^5 + 104^5 + 109^5 + 121^5,
so 55302546200 is in A345723, but is not in this sequence.

Examples

			9085584992 = 24^5 + 38^5 + 42^5 + 48^5 + 54^5 + 96^5
           = 21^5 + 34^5 + 38^5 + 43^5 + 74^5 + 92^5
           =  8^5 + 34^5 + 38^5 + 62^5 + 68^5 + 92^5
           = 18^5 + 18^5 + 44^5 + 64^5 + 66^5 + 92^5
           = 13^5 + 18^5 + 51^5 + 64^5 + 64^5 + 92^5
           =  8^5 + 38^5 + 41^5 + 47^5 + 79^5 + 89^5
           =  5^5 + 23^5 + 29^5 + 45^5 + 85^5 + 85^5
           =  8^5 + 23^5 + 41^5 + 64^5 + 82^5 + 84^5
           = 12^5 + 18^5 + 38^5 + 72^5 + 78^5 + 84^5,
so 9085584992 is a term.

Links

Crossrefs

Cf. A345723, A345821, A346286, A346363, A346365.

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 == 9])
    for x in range(len(rets)):
        print(rets[x])

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

Content is available under The OEIS End-User License Agreement.