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.

A324298 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 6 (A324297).

Original entry on oeis.org

3, 9, 15, 21, 25, 27, 33, 39, 41, 45, 51, 57, 63, 67, 69, 73, 75, 81, 87, 89, 93, 99, 105, 111, 117, 119, 121, 123, 129, 135, 137, 141, 145, 147, 153, 159, 165, 169, 171, 177, 183, 185, 189, 195, 197, 201, 207, 211, 213, 217, 219, 223, 225, 231, 233, 237, 243, 249
Offset: 1

Views

Author

Stefano Spezia, Mar 16 2019

Keywords

Comments

All the terms of this sequence are odd.
Why? If an integer 10*k+6 = (10*a+6) * (10*b+6), then k = 10*a*b + 6*(a+b) + 3, so k is odd. - Bernard Schott, May 13 2019

Examples

			145 is a term because 26*56 = 1456 = 145*10 + 6. - _Bernard Schott_, May 13 2019

Links

Crossrefs

Cf. A017341, A053742 (ending with 5), A324297, A337856, A346389.

Programs

  • Mathematica
    a={}; For[n=0,n<=250,n++,For[k=0,k<=n,k++,If[Mod[10*n+6,10*k+6]==0 && Mod[(10*n+6)/(10*k+6),10]==6 && 10*n+6>Max[10*a+6],AppendTo[a,n]]]]; a
  • PARI
    isok6(n) = (n%10) == 6; \\ A017341
isok(k) = {my(n=10*k+6, d=divisors(n)); fordiv(n, d, if (isok6(d) && isok6(n/d), return(1))); return (0);} \\ Michel Marcus, Apr 14 2019
  • Python
    def aupto(lim): return sorted(set(a*b//10 for a in range(6, 10*lim//6+2, 10) for b in range(a, 10*lim//a+2, 10) if a*b//10 <= lim))
print(aupto(249)) # Michael S. Branicky, Aug 21 2021

Formula

a(n) = (A324297(n) - 6)/10.
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 1.
The conjecture is true since a(n) = (A324297(n) - 6)/10 and lim_{n->infinity} A324297(n)/A324297(n-1) = 1. - Stefano Spezia, Aug 21 2021

A346388 a(n) is the number of proper divisors of A053742(n) ending with 5.

Original entry on oeis.org

1, 3, 2, 3, 5, 3, 3, 5, 3, 3, 7, 3, 3, 7, 3, 3, 7, 5, 3, 7, 3, 3, 8, 3, 5, 7, 3, 5, 7, 3, 3, 11, 5, 3, 7, 3, 3, 7, 7, 3, 9, 3, 5, 7, 3, 7, 7, 5, 3, 11, 3, 3, 11, 3, 3, 7, 3, 5, 11, 7, 5, 7, 4, 3, 7, 3, 7, 11, 3, 3, 7, 7, 5, 11, 3, 3, 11, 5, 3, 7, 7, 3, 11, 3, 5
Offset: 0

Views

Author

Stefano Spezia, Jul 15 2021

Keywords

Examples

			a(10) = 7 since there are 7 proper divisors of A053742(10) = 525 ending with 5: 5, 15, 25, 35, 75, 105 and 175.

Crossrefs

Cf. A032741, A053742, A346389 (ending with 6), A346392.

Programs

  • Mathematica
    a[n_]:=Length[Drop[Select[Divisors[25+50n], (Mod[#,10]==5&)], -1]]; Array[a, 90, 0]

Formula

a(n) = A346392(A053742(n)).

A346953 a(n) is the number of divisors of A346950(n) ending with 3.

Original entry on oeis.org

1, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 4, 4, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 2, 2, 4, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 2
Offset: 1

Views

Author

Stefano Spezia, Aug 08 2021

Keywords

Comments

a(n) = 1 if A346950(n) = k^2 where k is either a prime ending with 3 or the product of a prime ending with 7 and a prime ending with 9. - Robert Israel, Nov 03 2024

Examples

			a(17) = 4 since there are 4 divisors of A346950(17) = 429 ending with 3: 3, 13, 33 and 143.

Links

Crossrefs

Cf. A000005, A017377, A346388 (ending with 5), A346389 (ending with 6), A346950, A346951, A346952.

Programs

  • Maple
    N:= 10000: # for a(1) .. a(M) where the last term of A346950 less than N is A346950(M)
S:= {}:
for n from 3 to floor(sqrt(N)) by 10 do
  S:= S union map(`*`, {seq(i,i= n .. floor(N/n), 10)},n)
od:
S:= sort(convert(S,list)):
map(t -> nops(select(t -> t mod 10 = 3, numtheory:-divisors(t))), S); # Robert Israel, Nov 03 2024
  • Mathematica
    b={}; For[n=0, n<=450, n++, For[k=0, k<=n, k++, If[Mod[10*n+9, 10*k+3]==0 && Mod[(10*n+9)/(10*k+3), 10]==3 && 10*n+9>Max[b], AppendTo[b, 10*n+9]]]]; (* A346950 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==3&)]]]]]; a
  • Python
    from sympy import divisors
def f(n): return sum(d%10 == 3 for d in divisors(n)[1:-1])
def A346950upto(lim): return sorted(set(a*b for a in range(3, lim//3+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346950upto(2129)))) # Michael S. Branicky, Aug 11 2021

A346510 a(n) is the number of nontrivial divisors of A346507(n) ending with 1.

Original entry on oeis.org

1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 2, 4, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 4, 2, 2, 4, 2, 4, 2
Offset: 1

Views

Author

Stefano Spezia, Jul 21 2021

Keywords

Examples

			a(42) = 4 since there are 4 nontrivial divisors of A346507(42) = 2541 ending with 1: 11, 21, 121 and 231.

Crossrefs

Cf. A017281, A070824, A346388 (ending with 5), A346389 (ending with 6), A346392, A346507, A346508, A346509.

Programs

  • Mathematica
    b={}; For[n=1, n<=500, n++, For[k=1, kMax[b], AppendTo[b, 10n+1]]]]; (* A346507 *) a={}; For[i =1, i<=Length[b], i++, AppendTo[a, Length[Drop[Select[Divisors[Part[b, i]], (Mod[#, 10]==1&)], -1]]-1]]; a
  • PARI
    f(n) = sumdiv(n, d, (d>1) && (d(f(x)), [1..5000])) \\ Michel Marcus, Jul 28 2021
  • Python
    from sympy import divisors
def f(n): return sum(d%10 == 1 for d in divisors(n)[1:-1])
def A346507upto(lim): return sorted(set(a*b for a in range(11, lim//11+1, 10) for b in range(a, lim//a+1, 10)))
print(list(map(f, A346507upto(5000)))) # Michael S. Branicky, Jul 31 2021

Formula

a(n) = A346392(A346507(n)) - 1.

A346651 a(n) is the number of divisors of A139245(n) ending with 2.

Original entry on oeis.org

1, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 3, 2, 4, 2, 2, 3, 2, 3, 4, 2, 2, 3, 2, 3, 5, 2, 3, 3, 2, 2, 4, 3, 2, 3, 4, 2, 4, 2, 3, 3, 2, 2, 5, 3, 2, 6, 2, 2, 4, 2, 3, 3, 3, 2, 4, 2, 4, 3, 3, 3, 5, 2, 2, 3, 2, 2, 7, 4, 2, 4, 2, 2, 4, 3, 3, 3, 2, 3, 6, 2, 3, 3, 4, 2, 4, 2, 2, 5, 2
Offset: 1

Views

Author

Stefano Spezia, Jul 26 2021

Keywords

Comments

a(n) is odd if and only if A139245(n) either is the square of a number ending with 2 or has a unitary prime divisor ending with 7.
The term 1 appears only for n = 1 in corresponding to A139245(1) = 4.

Examples

			a(14) = 4 since there are 4 divisors of A139245(14) = 264 ending with 2: 2, 12, 22 and 132.

Crossrefs

Cf. A000005, A017293 (numbers ending with 2), A017294 (squares of numbers ending with 2), A030432, A056169, A139245 (product of two numbers ending with 2), A346388, A346389.

Programs

  • Mathematica
    a[n_]:=Length[Drop[Select[Divisors[20n-16], (Last[IntegerDigits[#]]==2&)]]]; Array[a, 90]
  • PARI
    a(n) = sumdiv(20*n-16, d, (d%10) == 2); \\ Michel Marcus, Jul 26 2021
Showing 1-5 of 5 results.

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