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 1-8 of 8 results.

A346950 Positive integers k that are the product of two integers ending with 3.

Original entry on oeis.org

9, 39, 69, 99, 129, 159, 169, 189, 219, 249, 279, 299, 309, 339, 369, 399, 429, 459, 489, 519, 529, 549, 559, 579, 609, 639, 669, 689, 699, 729, 759, 789, 819, 849, 879, 909, 939, 949, 969, 989, 999, 1029, 1059, 1079, 1089, 1119, 1149, 1179, 1209, 1219, 1239, 1269
Offset: 1

Views

Author

Stefano Spezia, Aug 08 2021

Keywords

Comments

All the terms end with 9 (A017377).

Examples

			9 = 3*3, 39 = 3*13, 69 = 3*23, 99 = 3*33, 129 = 3*43, 159 = 3*53, 169 = 13*13, 189 = 3*63, ...

Links

Crossrefs

Cf. A017377 (supersequence), A053742 (ending with 5), A139245 (ending with 2), A324297 (ending with 6), A346951, A346952, A346953.

Programs

  • Mathematica
    a={}; For[n=0, n<=250, 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[a], AppendTo[a, 10*n+9]]]]; a
  • Python
    def aupto(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(aupto(1270)) # Michael S. Branicky, Aug 08 2021

Formula

Limit_{n->oo} a(n)/a(n-1) = 1.

A347255 Number of positive integers with n digits that are the product of two integers ending with 4.

Original entry on oeis.org

0, 3, 25, 281, 2941, 30596, 315385, 3231664, 32972224, 335346193, 3402373313, 34454358909, 348373701706, 3518101287286, 35491654274101
Offset: 1

Views

Author

Stefano Spezia, Aug 24 2021

Keywords

Comments

a(n) is the number of n-digit numbers in A347253.

Crossrefs

Cf. A017341, A052268, A347253.
Cf. A346509 (ending with 1), A346952 (ending with 3), A337855 (ending with 5), A337856 (ending with 6).

Programs

  • Python
    def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(4, hi//4+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Aug 24 2021

Formula

a(n) < A052268(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

Extensions

a(9)-a(11) from Michael S. Branicky, Aug 25 2021
a(12)-a(15) from Martin Ehrenstein, Sep 29 2021

A348055 Number of positive integers with n digits that are the product of two integers ending with 7.

Original entry on oeis.org

0, 1, 20, 255, 3064, 34743, 380939, 4089499, 43282317, 453472867, 4715695283, 48760330737, 501941505404, 5148657883067, 52659616820819
Offset: 1

Views

Author

Stefano Spezia, Sep 26 2021

Keywords

Comments

a(n) is the number of n-digit numbers in A348054.

Crossrefs

Cf. A017377, A052268, A348054.
Cf. A346509 (ending with 1), A346629 (ending with 2), A346952 (ending with 3), A347255 (ending with 4), A337855 (ending with 5), A337856 (ending with 6), A348549 (ending with 8).

Programs

  • Python
    def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(7, hi//7+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Sep 26 2021

Formula

a(n) < A052268(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

Extensions

a(9)-a(11) from Michael S. Branicky, Sep 26 2021
a(12)-a(15) from Martin Ehrenstein, Oct 25 2021

A346951 Positive integers k such that 10*k+9 is equal to the product of two integers ending with 3 (A346950).

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 16, 18, 21, 24, 27, 29, 30, 33, 36, 39, 42, 45, 48, 51, 52, 54, 55, 57, 60, 63, 66, 68, 69, 72, 75, 78, 81, 84, 87, 90, 93, 94, 96, 98, 99, 102, 105, 107, 108, 111, 114, 117, 120, 121, 123, 126, 129, 132, 133, 135, 138, 141, 144, 146, 147, 150
Offset: 1

Views

Author

Stefano Spezia, Aug 08 2021

Keywords

Examples

			15 is a term because 3*53 = 159 = 15*10 + 9.

Crossrefs

Cf. A016873 (ending with 5), A017377, A324298 (ending with 6), A346950, A346952, A346953.

Programs

  • Mathematica
    a={}; For[n=0, n<=150, 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[10a+9], AppendTo[a, n]]]]; a
  • Python
    def aupto(lim): return sorted(set(a*b//10 for a in range(3, 10*lim//3+4, 10) for b in range(a, 10*lim//a+4, 10) if a*b//10 <= lim))
print(aupto(150)) # Michael S. Branicky, Aug 11 2021

Formula

a(n) = (A346950(n) - 9)/10.
Lim_{n->infinity} a(n)/a(n-1) = 1.

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

A348546 Number of positive integers with n digits that are equal both to the product of two integers ending with 3 and to that of two integers ending with 7.

Original entry on oeis.org

0, 0, 8, 129, 1771, 21802, 252793, 2826973, 30872783
Offset: 1

Views

Author

Stefano Spezia, Oct 22 2021

Keywords

Comments

a(n) is the number of n-digit numbers in A348544.

Crossrefs

Cf. A052268, A346950, A346952, A348054, A348055, A348544, A348547.

Programs

  • Mathematica
    Table[{lo, hi}={10^(n-1), 10^n}; Length@Select[Intersection[Union@Flatten@Table[a*b, {a, 3, Floor[hi/3], 10}, {b, a, Floor[hi/a], 10}], Union@Flatten@Table[a*b, {a, 7, Floor[hi/7], 10}, {b, a, Floor[hi/a], 10}]], lo<#
  • Python
    def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(3, hi//3+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi) & set(a*b for a in range(7, hi//7+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Oct 22 2021

Formula

a(n) < A052268(n).
a(n) = A346952(n) + A348055(n) - A348547(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

Extensions

a(9) from Michael S. Branicky, Oct 22 2021

A348547 Number of positive integers with n digits and final digit 9 that are equal to the product of two integers ending with the same digit.

Original entry on oeis.org

1, 4, 49, 524, 5596, 58706, 608886, 6267854, 64180304, 654605898, 6656849267, 67539297095, 683989985496, 6916722312963, 69859080168037
Offset: 1

Views

Author

Stefano Spezia, Oct 22 2021

Keywords

Comments

a(n) is the number of n-digit numbers in A348545.

Crossrefs

Cf. A052268, A346950, A346952, A348054, A348055, A348545, A348546.

Programs

  • Mathematica
    Table[{lo, hi}={10^(n-1), 10^n}; Length@Select[Union[Union@Flatten@Table[a*b, {a, 3, Floor[hi/3], 10}, {b, a, Floor[hi/a], 10}], Union@Flatten@Table[a*b, {a, 7, Floor[hi/7], 10}, {b, a, Floor[hi/a], 10}]], lo<#
  • Python
    def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(3, hi//3+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi) | set(a*b for a in range(7, hi//7+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Oct 22 2021

Formula

a(n) < A052268(n).
a(n) = A346952(n) + A348055(n) - A348546(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

Extensions

a(9) from Michael S. Branicky, Oct 22 2021
a(10)-a(15) from Martin Ehrenstein, Nov 06 2021

A348549 Number of positive integers with n digits that are the product of two integers ending with 8.

Original entry on oeis.org

0, 1, 14, 195, 2200, 24013, 255969, 2687317, 27934809, 288342379, 2960920297, 30285890402, 308834717932, 3141625339760, 31895159990436
Offset: 1

Views

Author

Stefano Spezia, Oct 22 2021

Keywords

Comments

a(n) is the number of n-digit numbers in A348548.

Crossrefs

Cf. A052268, A348548.
Cf. A346509 (ending with 1), A346629 (ending with 2), A346952 (ending with 3), A347255 (ending with 4), A337855 (ending with 5), A337856 (ending with 6), A348055 (ending with 7).

Programs

  • Python
    def a(n):
  lo, hi = 10**(n-1), 10**n
  return len(set(a*b for a in range(8, hi//8+1, 10) for b in range(a, hi//a+1, 10) if lo <= a*b < hi))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Oct 22 2021

Formula

a(n) < A052268(n).
Conjecture: lim_{n->infinity} a(n)/a(n-1) = 10.

Extensions

a(9)-a(10) from Michael S. Branicky, Oct 22 2021
a(11)-a(15) from Martin Ehrenstein, Nov 06 2021
Showing 1-8 of 8 results.

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