A346951 -id:A346951 - cp's OEIS frontend

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

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

Stefano Spezia, Aug 08 2021

All the terms end with 9 (A017377).

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

Stefano Spezia, Table of n, a(n) for n = 1..10000

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

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

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

Limit_{n->oo} 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

Stefano Spezia, Aug 08 2021

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

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

Robert Israel, Table of n, a(n) for n = 1..10000

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

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

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

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

A347254 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 4 (A347253).

Original entry on oeis.org

1, 5, 9, 13, 17, 19, 21, 25, 29, 33, 37, 41, 45, 47, 49, 53, 57, 61, 65, 69, 73, 75, 77, 81, 85, 89, 93, 97, 101, 103, 105, 109, 113, 115, 117, 121, 125, 129, 131, 133, 137, 141, 145, 149, 153, 157, 159, 161, 165, 169, 173, 177, 181, 183, 185, 187, 189, 193, 197

Offset: 1

Stefano Spezia, Aug 24 2021

Since an integer 10*k + 6 = (10*a + 4)*(10*b + 4) implies that k = 10*a*b + 4*(a + b) + 1, all the terms of this sequence are odd.

			13 is a term because 4*34 = 136 = 13*10 + 6.

Cf. A016873 (ending with 5), A017341, A324298 (ending with 6), A346951 (ending with 3), A347253.

a={}; For[n=0, n<=200, n++, For[k=0, k<=n, k++, If[Mod[10*n+6, 10*k+4]==0 && Mod[(10*n+6)/(10*k+4), 10]==4 && n>Max[a], AppendTo[a, n]]]]; a

isok(k) =  my(x=10*k+6); sumdiv(x, d, (Mod(d, 10)==4) && Mod(x/d, 10)==4); \\ Michel Marcus, Oct 04 2021

def aupto(lim): return sorted(set(a*b//10 for a in range(4, 10*lim//4+3, 10) for b in range(a, 10*lim//a+3, 10) if a*b//10 <= lim))
print(aupto(197)) # Michael S. Branicky, Aug 24 2021

a(n) = (A347253(n) - 6)/10.

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

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

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

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A347254 Positive integers k such that 10*k+6 is equal to the product of two integers ending with 4 (A347253).

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