A358270 Numbers whose sum of digits is even and that have an even number of even digits.

11, 13, 15, 17, 19, 20, 22, 24, 26, 28, 31, 33, 35, 37, 39, 40, 42, 44, 46, 48, 51, 53, 55, 57, 59, 60, 62, 64, 66, 68, 71, 73, 75, 77, 79, 80, 82, 84, 86, 88, 91, 93, 95, 97, 99, 1001, 1003, 1005, 1007, 1009, 1010, 1012, 1014, 1016, 1018, 1021, 1023, 1025, 1027, 1029, 1030

Offset: 1

Author

Bernard Schott, Nov 06 2022

Keywords

• nonn
• base
• easy

Comments

There are only terms with an even number of digits, and precisely, there exist A137233(2*k) terms with 2*k digits.

The conditions separately are A054683 for even sum of digits, and A356929 for even number of even digits, so that this sequence is their intersection.

The opposite conditions, an odd sum of digits, and an odd number of odd digits, are the same and are A054684.

Examples

			26 is a term since 2+6 = 8 (even) and 26 has two even digits.
39 is a term since 3+9 = 12 (even) and 39 has zero even digits.
1012 is a term since 1+0+1+2 = 4 (even) and 1012 has two even digits.
		

Crossrefs

Intersection of A054683 and A356929.

Cf. A001637 (even length), A179081 (digit sum mod 2).

Cf. A014263, A054684, A137233.

Programs

Mathematica

Select[Range[1000], EvenQ[Plus @@ IntegerDigits[#]] && EvenQ[Plus @@ DigitCount[#, 10, Range[0, 8, 2]]] &] (* Amiram Eldar, Nov 06 2022 *)

PARI

a(n) = n*=2; n += 100^logint(110*n,100) \ 11; n - sumdigits(n)%2; \\ Kevin Ryde, Nov 10 2022

Python

def ok(n): s = str(n); return sum(map(int, s))%2 == sum(1 for d in s if d in "02468")%2 == 0
print([k for k in range(1031) if ok(k)]) # Michael S. Branicky, Nov 06 2022

Python

from itertools import count, islice, chain
def A358270_gen(): # generator of terms
    return filter(lambda n:not (len(s:=str(n))&1 or sum(int(d) for d in s)&1), chain.from_iterable((range(10**l,10**(l+1)) for l in count(1,2))))
A358270_list = list(islice(A358270_gen(),61)) # Chai Wah Wu, Nov 11 2022

Formula

a(n) = t - A179081(t) where t = A001637(2*n). - Kevin Ryde, Nov 10 2022