A359271
Number of odd digits necessary to write all nonnegative n-digit integers.
Original entry on oeis.org
5, 95, 1400, 18500, 230000, 2750000, 32000000, 365000000, 4100000000, 45500000000, 500000000000, 5450000000000, 59000000000000, 635000000000000, 6800000000000000, 72500000000000000, 770000000000000000, 8150000000000000000
Offset: 1
To write the integers from 10 up to 99, each of the digits 1, 3, 5, 7 and 9, must be used 19 times, hence a(2) = 19*5 = 95.
-
seq(5 * (9*n+1) * 10^(n-2), n=1..18);
-
a[n_] := 5*(9*n + 1)*10^(n - 2); Array[a, 20] (* Amiram Eldar, Dec 23 2022 *)
A386475
Least prime starting a run of exactly n consecutive primes with identical counts of odd digits.
Original entry on oeis.org
2, 5, 3, 11, 97, 503, 499, 491, 14303, 14293, 157259, 157253, 1525723, 4576997, 4576993, 4576991, 10411013, 33388093, 188332121, 194259301, 2853982501, 2853982499, 2853982477, 3913474277, 10883385143, 22809734971, 34883348389, 34883348369, 34883348341
Offset: 1
a(2) = 5, because the two primes in the sequence starting at 5, namely 5 and 7, each contain the same number of odd digits, and no earlier prime sequence meets this criterion.
In [2], each number contains 0 odd digits.
In [5, 7], each number contains 1 odd digit.
In [3, 5, 7], each number contains 1 odd digit.
In [11, 13, 17, 19], each number contains 2 odd digits.
In [97, 101, 103, 107, 109], each number contains 2 odd digits.
In [503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [499, 503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [491, 499, 503, 509, 521, 523, 541, 547], each number contains 2 odd digits.
In [14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each number contains 3 odd digits.
In [14293, 14303, 14321, 14323, 14327, 14341, 14347, 14369, 14387, 14389], each number contains 3 odd digits.
-
oddn[n_] := Plus @@ Mod[IntegerDigits@ n, 2]; T = Table[0, {99}]; p = 1; While[p < 2 10^6, p = NextPrime[p]; c = oddn[p]; r=1; q=p; While[True, q = NextPrime[q]; If[oddn[q] == c, r++, Break[]]]; If[T[[r]] == 0, T[[r]] = p]]; Take[T, Position[T, 0][[1, 1]] - 1] (* Giovanni Resta, Jul 23 2025 *)
A352751
Modified Sisyphus function of order 4: a(n) is the concatenation of (number of digits of n)(number digits of n congruent to 0 modulo 4)(number of digits of n congruent to 1 modulo 4)(number of digits of n congruent to 2 modulo 4)(number of digits of n congruent to 3 modulo 4).
Original entry on oeis.org
11000, 10100, 10010, 10001, 11000, 10100, 10010, 10001, 11000, 10100, 21100, 20200, 20110, 20101, 21100, 20200, 20110, 20101, 21100, 20200, 21010, 20110, 20020, 20011, 21010, 20110, 20020, 20011, 21010, 20110, 21001, 20101, 20011, 20002, 21001, 20101, 20011, 20002, 21001, 20101, 22000, 21100, 21010
Offset: 0
11 has two digits, both congruent to 1 modulo 4, so a(11) = 20200.
a(20) = 21010.
a(30) = 21001.
a(1111123567) = 100622.
- M. E. Coppenbarger, Iterations of a modified Sisyphus function, Fib. Q., 56 (No. 2, 2018), 130-141.
-
def a(n, order=4):
d, m = list(map(int, str(n))), [0]*order
for di in d: m[di%order] += 1
return int(str(len(d)) + "".join(map(str, m)))
print([a(n) for n in range(37)]) # Michael S. Branicky, Apr 01 2022
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