A346509
Number of positive integers with n digits that are the product of two integers greater than 1 and ending with 1.
Original entry on oeis.org
0, 0, 12, 200, 2660, 31850, 361985, 3982799, 42914655, 455727689, 4788989458, 49930700093, 517443017072, 5336861879564
Offset: 1
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a(n) = {my(res = 0); forstep(i = 10^(n-1) + 1, 10^n, 10, f = factor(i); if(bigomega(f) == 1, next); d = divisors(f); for(j = 2, (#d~ + 1)>>1, if(d[j]%10 == 1 && d[#d + 1 - j]%10 == 1, res++; next(2) ) ) ); res } \\ David A. Corneth, Jul 22 2021
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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)))
def a(n): return len(A346507upto(10**n)) - len(A346507upto(10**(n-1)))
print([a(n) for n in range(1, 9)]) # Michael S. Branicky, Jul 22 2021
A073505
Number of primes == 1 (mod 10) less than 10^n.
Original entry on oeis.org
0, 5, 40, 306, 2387, 19617, 166104, 1440298, 12711386, 113761519, 1029517130, 9401960980, 86516370000
Offset: 1
a(2) = 5 because there are 5 primes == 1 (mod 10) less than 10^2. They are 11, 31, 41, 61 and 71.
Cf.
A006880,
A087630,
A073506,
A073507,
A073508,
A073509,
A073510,
A073511,
A073512,
A073513,
A073514,
A073515,
A073516,
A073517.
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c = 0; k = 1; Do[While[k < 10^n, If[PrimeQ[k], c++ ]; k += 10]; Print[c], {n, 1, 10}]
A087631
Number of n-digit primes ending in 3 in base 10.
Original entry on oeis.org
1, 6, 35, 268, 2092, 17263, 146565, 1274244, 11272025, 101053126, 915743823, 8372470456, 77114448042
Offset: 1
Meenakshi Srikanth (menakan_s(AT)yahoo.com) and Amarnath Murthy, Sep 15 2003
a(2) = 6, as there exist 6 two-digit prime numbers (13, 23, 43, 53, 73, and 83) with units place 3.
a(3) = 35, since there are 35 three-digit numbers with units place digit as 3.
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/** The terms of the sequences are generated by changing the range for j for the various numbers of digits. E.g., it ranges from 100 to 999 for three-digit numbers. */
float r, x;
int c = 0, count = 0;
for (float j = 100f; j < 1000f; j++) { for (float i = 2f; i < j; i++) { r = j % i; if (r == 0) c = 1; } if (c == 0) { x = j % 10; if (x == 3) count = count + 1; } c = 0; } System.out.println("count = " + count);
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Table[Length[Select[Range[10^n + 3, 10^(n + 1) - 7, 10], PrimeQ[#] &]], {n, 5}] (* Alonso del Arte, Apr 27 2014 *)
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a(n) = my(c=0); forprime(p=10^(n-1), 10^n, if(p%10==3, c++)); c \\ Iain Fox, Aug 07 2018
Offset corrected by
Iain Fox, Aug 07 2018
A087632
Number of n-digit primes ending in 7 in base 10.
Original entry on oeis.org
1, 5, 40, 262, 2103, 17210, 146590, 1274284, 11271819, 101051725, 915754298, 8372478663, 77114370790
Offset: 1
Meenakshi Srikanth (menakan_s(AT)yahoo.com) and Amarnath Murthy, Sep 15 2003
a(2) = 5 as there exist 5 two-digit prime numbers (17, 37, 47, 67, and 97) with units place 7.
a(3) = 40, since there are 40 three-digit numbers with units place digit as 7.
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/** The terms of the sequences are generated by changing the range for j for the various numbers of digits. E.g., it ranges from 100 to 999 for three-digit numbers. */
float r, x;
int c = 0, count = 0;
for (float j = 100f; j < 1000f; j++) { for (float i = 2f; i < j; i++) { r = j % i; if (r == 0) c = 1; } if (c == 0) { x = j % 10; if (x == 7) count = count + 1; } c = 0; } System.out.println("count = " + count);
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Table[Length[Select[Range[10^n + 7, 10^(n + 1) - 3, 10], PrimeQ[#] &]], {n, 5}] (* Alonso del Arte, Apr 27 2014 *)
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a(n) = my(c=0); forprime(p=10^(n-1), 10^n, if(p%10==7, c++)); c \\ Iain Fox, Aug 07 2018
Offset corrected by
Iain Fox, Aug 07 2018
A087633
Number of n-digit primes ending in 9 in base 10.
Original entry on oeis.org
0, 5, 33, 265, 2087, 17203, 146439, 1274154, 11271147, 101049993, 915748570, 8372464236, 77114396969
Offset: 1
Meenakshi Srikanth (menakan_s(AT)yahoo.com) and Amarnath Murthy, Sep 15 2003
a(2) = 5 as there exist 5 two-digit prime numbers (19, 29, 59, 79, and 89) with units place 9.
a(3) = 33, since there are 33 three-digit numbers with units place digit as 9.
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/** The terms of the sequences are generated by changing the range for j for the various numbers of digits. E.g., it ranges from 100 to 999 for three-digit numbers. */
float r, x;
int c = 0, count = 0; for (float j = 100f; j < 1000f; j++) { for (float i = 2f; i < j; i++) { r = j % i; if (r == 0) c = 1; } if (c == 0) { x = j % 10; if (x == 9) count = count + 1; } c = 0; } System.out.println("count = " + count);
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Table[Length[Select[Range[10^n + 9, 10^(n + 1) - 1, 10], PrimeQ]], {n, 5}] (* Alonso del Arte, Apr 27 2014 *)
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a(n) = my(c=0); forprime(p=10^(n-1), 10^n, if(p%10==9, c++)); c \\ Iain Fox, Aug 07 2018
Offset corrected by
Iain Fox, Aug 07 2018
Showing 1-5 of 5 results.
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